Question

DX/x+y=dy/x+y=DX/-(x+y+2z)

Answer 1

Given, $\dfrac{dx}{x+y}=\dfrac{dy}{x+y}=\dfrac{dx}{-(x+y+2z)}$

Taking first two terms-

$\dfrac{dx}{x+y}=\dfrac{dy}{x+y}$

$\Rightarrow dx=dy$

Integrate-

$x=y+c_1$

$c_1=x-y~~~-(1)$

Taking first and last term we get-

$\dfrac{dx}{x+y}=\dfrac{dx}{-(x+y+2z)} \\[9pt] \Rightarrow x+y=-x-y-2z \\[9pt] \Rightarrow z=-(x+y)$

Now Taking second and third term-

$\dfrac{dy}{x+y}=\dfrac{dz}{-(x+y+2(-x-y)} \\[9pt] \Rightarrow \dfrac{dy}{x+y}=\dfrac{dx}{x+y} \\[9pt] \Rightarrow dy=dx$

Integrate-

$y=x+c_2\\[9pt] c_2=y-x~~~~~-(2)$

The solution is-

$\phi(c_1,c_2)=0 \\[9pt] \phi(x-y,y-x)=0$